Y less than x minus 4

Given the pre-image ABCD and the image after a dilation, A'B'C'D', what is true about the polygons? Check all that apply. The le

vazorg [7]

<h2>Answer </h2>

You should check the following:

The length of AB is 2

The length of A'B' is 3

The scale factor is $\frac{3}{2}$

<h2>Explanation</h2>

- To find the length of AB, we will use the distance formula:

$d(A,B)=|x_{1}-x_{2}|$

From the picture we can infer that $x_{1}=2$ and $x_{2}=4$ , so let's replace the values.

$d(A,B)=|2-4|$

$d(A,B)=|-2|$

$d(A,B)=2$

The length of AB is 2, so you should check the first answer

- To find the length of A'B', we will use the distance formula one more time:

From the picture we can infer that $x_{1}=3$ and $x_{2}=6$ , so let's replace the values.

$d(A',B')=|3-6|$

$d(A',B')=|-3|$

$d(A',B')=3$

Since the length of A'B' is 3, you should also check the second answer as well.

- In a transformation (in this case a dilation) the pre-image is the original shape and the image is the shape after the transformation.

Since the image is bigger than the pre-image, the third answer is false.

- Let $k$ be the scale factor

We know that the length of AB is 2 and we need to multiply that length by the scale factor to get length of A'B' i.e 3, so:

$2k=3$

$k=\frac{3}{2}$

Since the scale factor is $\frac{3}{2}$ , you should check the fourth answer as well.

- We already prove that the scale factor is $\frac{3}{2}$ , so the fifth answer is false.

5 0

4 years ago

Read 2 more answers

Suppose that during a test drive of two cars, one car travels 186 miles in the same time that the second car travels 150 miles.

nevsk [136]

<span><span>Algebra: Exponent and logarithm as functions of powerSolversLessonsAnswers archive </span><span>

<span>Click here to see ALL problems on Exponential-and-logarithmic-functions<span>Question 726619: <span>Skid mark analysis is one way to estimate the speed a car was traveling prior to an accident. The speed, s, that the vehicle was traveling in miles per hour can be approximated by the formula s=square root 30df, where d represents the length of the skid marks in fee and "f" represents the drag factor of the road. A vehicle that was involved in an accident on an asphalt road was traveling at a speed of 45 mph when it started skidding. If the drag factor for asphalt is 0.75 find the length of the skid marks made by the car. The skid marks made by the car were approximately___________ feet long. </span>
<span>Answer by </span>stanbon(75310) (Show Source):You can put this solution on YOUR website!
<span>s=square root 30df, where d represents the length of the skid marks in feet and "f" represents the drag factor of the road. A vehicle that was involved in an accident on an asphalt road was traveling at a speed of 45 mph when it started skidding. If the drag factor for asphalt is 0.75 find the length of the skid marks made by the car.
S = sqrt(30df)
45 = sqrt(30*d*0.75)
----
45 = sqrt[(90/4)d]
-----
45 = (3/2)sqrt(10d)
30 = sqrt(10d)
10d = 900
d = 90 ft.
================
</span></span></span></span></span>

5 0

3 years ago

Which statement is not true by CPCTC

Nana76 [90]

The answer to your question is the third statement

5 0

3 years ago

Read 2 more answers

Which of the following is the reason for statement #7 in the table above

Rudiy27

There is no table above.

Also there are no choices of answers.

3 0

4 years ago

If a van traveled 200 miles in 5hours how far could the van go in 8hours

enyata [817]

Answer: The Answer should be 320 miles in 8 hours.

Step-by-step explanation: Well 200 divided by 5 is 40 so that's how many miles they travel in one hour so all you have to do is multiply 8 x 40 which would give you the answer of 320 miles in 8 hours.

7 0

3 years ago

Read 2 more answers